How To Change From Polar To Rectangular

10.3: Polar Coordinates

1. Concluding updated

2. Save as PDF

Page ID

2682

Learning Objectives

• Plot points using polar coordinates.
• Catechumen from polar coordinates to rectangular coordinates.
• Convert from rectangular coordinates to polar coordinates.
• Transform equations between polar and rectangular forms.
• Place and graph polar equations by converting to rectangular equations.

Over \(12\) kilometers from port, a sailboat encounters rough weather and is diddled off class by a \(sixteen\)-knot wind (meet Figure \(\PageIndex{one}\)). How can the sailor point his location to the Coast Guard? In this section, nosotros will investigate a method of representing location that is different from a standard coordinate grid.

Effigy \(\PageIndex{1}\)

Plotting Points Using Polar Coordinates

When we think nearly plotting points in the plane, nosotros unremarkably recollect of rectangular coordinates \((x,y)\) in the Cartesian coordinate airplane. Even so, there are other ways of writing a coordinate pair and other types of filigree systems. In this department, we introduce to polar coordinates, which are points labeled \((r,\theta)\) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane.

The polar filigree is scaled as the unit circle with the positive \(ten\)-axis now viewed equally the polar axis and the origin as the pole. The first coordinate \(r\) is the radius or length of the directed line segment from the pole. The angle \(\theta\), measured in radians, indicates the direction of \(r\). Nosotros move counterclockwise from the polar axis by an angle of \(\theta\),and measure a directed line segment the length of \(r\) in the direction of \(\theta\). Even though we measure \(\theta\) first and then \(r\), the polar point is written with the \(r\)-coordinate showtime. For example, to plot the point \(\left(2,\dfrac{\pi}{4}\right)\),we would motility \(\dfrac{\pi}{4}\) units in the counterclockwise management and then a length of \(2\) from the pole. This bespeak is plotted on the grid in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\)

Example \(\PageIndex{1}\): Plotting a Betoken on the Polar Filigree

Plot the betoken \(\left(3,\dfrac{\pi}{two}\right)\) on the polar grid.

Solution

The angle \(\dfrac{\pi}{two}\) is found by sweeping in a counterclockwise direction \(90°\) from the polar axis. The bespeak is located at a length of \(three\) units from the pole in the \(\dfrac{\pi}{2}\) direction, every bit shown in Figure \(\PageIndex{three}\).

Figure \(\PageIndex{3}\)

Exercise \(\PageIndex{ane}\)

Plot the bespeak \(\left(two, \dfrac{\pi}{3}\right)\) in the polar grid.

Answer

Figure \(\PageIndex{4}\)

Example \(\PageIndex{2}\): Plotting a Point in the Polar Coordinate System with a Negative Component

Plot the point \(\left(−2, \dfrac{\pi}{6}\right)\) on the polar grid.

Solution

We know that \(\dfrac{\pi}{6}\) is located in the outset quadrant. However, \(r=−2\). Nosotros can approach plotting a point with a negative \(r\) in ii ways:

1. Plot the point \(\left(2,\dfrac{\pi}{6}\right)\) by moving \(\dfrac{\pi}{6}\) in the counterclockwise management and extending a directed line segment \(2\) units into the first quadrant. Then retrace the directed line segment back through the pole, and proceed \(2\) units into the third quadrant;
2. Movement \(\dfrac{\pi}{6}\) in the counterclockwise direction, and draw the directed line segment from the pole \(2\) units in the negative direction, into the third quadrant.

See Effigy \(\PageIndex{5a}\). Compare this to the graph of the polar coordinate \((ii,π6)\) shown in Effigy \(\PageIndex{5b}\).

Figure \(\PageIndex{5}\)

Practice \(\PageIndex{2}\)

Plot the points \(\left(3,−\dfrac{\pi}{6}\right)\) and \(\left(two,\dfrac{9\pi}{four}\correct)\) on the same polar filigree.

Answer

Figure \(\PageIndex{vi}\)

Converting from Polar Coordinates to Rectangular Coordinates

When given a ready of polar coordinates, we may need to convert them to rectangular coordinates. To exercise and so, we tin can think the relationships that exist amidst the variables \(ten\), \(y\), \(r\), and \(\theta\).

\(\cos \theta=\dfrac{ten}{r}\rightarrow x=r \cos \theta\)

\(\sin \theta=\dfrac{y}{r}\rightarrow y=r \sin \theta\)

Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in Figure \(\PageIndex{7}\). An easy manner to call back the equations above is to remember of \(\cos \theta\) equally the adjacent side over the hypotenuse and \(\sin \theta\) as the contrary side over the hypotenuse.

Figure \(\PageIndex{7}\)

CONVERTING FROM POLAR COORDINATES TO RECTANGULAR COORDINATES

To convert polar coordinates \((r, \theta)\) to rectangular coordinates \((ten, y)\), allow

\[\cos \theta=\dfrac{x}{r}\rightarrow x=r \cos \theta\]

\[\sin \theta=\dfrac{y}{r}\rightarrow y=r \sin \theta\]

How to: Given polar coordinates, convert to rectangular coordinates.

1. Given the polar coordinate \((r,\theta)\), write \(x=r \cos \theta\) and \(y=r \sin \theta\).
2. Evaluate \(\cos \theta\) and \(\sin \theta\).
3. Multiply \(\cos \theta\) by \(r\) to find the \(x\)-coordinate of the rectangular grade.
4. Multiply \(\sin \theta\) by \(r\) to notice the \(y\)-coordinate of the rectangular form.

Instance \(\PageIndex{3A}\): Writing Polar Coordinates equally Rectangular Coordinates

Write the polar coordinates \(\left(iii,\dfrac{\pi}{2}\right)\) as rectangular coordinates.

Solution

Use the equivalent relationships.

\[\brainstorm{align*} x&= r \cos \theta\\ x&= 3 \cos \dfrac{\pi}{ii}\\ &= 0\\ y&= r \sin \theta\\ y&= 3 \sin \dfrac{\pi}{2}\\ &= three \end{marshal*}\]

The rectangular coordinates are \((0,3)\). See Effigy \(\PageIndex{8}\).

Figure \(\PageIndex{eight}\)

Instance \(\PageIndex{3B}\): Writing Polar Coordinates as Rectangular Coordinates

Write the polar coordinates \((−2,0)\) as rectangular coordinates.

Solution

See Figure \(\PageIndex{ix}\). Writing the polar coordinates every bit rectangular, nosotros have

\[\begin{marshal*} x&= r \cos \theta\\ x&= -2 \cos(0)\\ &= -two\\ y&= r \sin \theta\\ y&= -two \sin(0)\\ &= 0 \finish{align*}\]

The rectangular coordinates are also \((−2,0)\).

Figure \(\PageIndex{ix}\)

Exercise \(\PageIndex{three}\)

Write the polar coordinates \(\left(−1,\dfrac{2\pi}{3}\right)\) equally rectangular coordinates.

Answer

\((ten,y)=\left(\dfrac{1}{2},−\dfrac{\sqrt{3}}{2}\right)\)

Converting from Rectangular Coordinates to Polar Coordinates

To convert rectangular coordinates to polar coordinates, we will utilise two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates volition yield more than than one polar signal.

CONVERTING FROM RECTANGULAR COORDINATES TO POLAR COORDINATES

Converting from rectangular coordinates to polar coordinates requires the use of one or more than of the relationships illustrated in Figure \(\PageIndex{10}\).

\(\cos \theta=\dfrac{x}{r}\) or \(x=r \cos \theta\)

\(\sin \theta=\dfrac{y}{r}\) or \(y=r \sin \theta\)

\(r^2=x^2+y^2\)

\(\tan \theta=\dfrac{y}{x}\)

Effigy \(\PageIndex{10}\)

Instance \(\PageIndex{4}\): Writing Rectangular Coordinates as Polar Coordinates

Convert the rectangular coordinates \((3,3)\) to polar coordinates.

Solution

Nosotros run into that the original signal \((3,iii)\) is in the first quadrant. To find \(\theta\), utilise the formula \(\tan \theta=\dfrac{y}{x}\). This gives

\[\begin{align*} \tan \theta&= \dfrac{3}{3}\\ \tan \theta&= one\\ {\tan}^{-1}(1)&= \dfrac{\pi}{4} \end{align*}\]

To find \(r\), we substitute the values for \(x\) and \(y\) into the formula \(r=\sqrt{x^2+y^two}\). We know that \(r\) must be positive, equally \(\dfrac{\pi}{iv}\) is in the first quadrant. Thus

\[\brainstorm{align*} r&= \sqrt{3^ii+iii^two}\\ r&= \sqrt{9+9}\\ r&= \sqrt{18}\\ &= 3\sqrt{2} \end{align*}\]

So, \(r=iii\sqrt{2}\) and \(\theta=\dfrac{\pi}{4}\), giving us the polar signal \((3\sqrt{2},\dfrac{\pi}{four})\). See Figure \(\PageIndex{11}\).

Figure \(\PageIndex{11}\)

Analysis

At that place are other sets of polar coordinates that will be the aforementioned every bit our first solution. For example, the points \(\left(−three\sqrt{2}, \dfrac{v\pi}{4}\right)\) and \(\left(3\sqrt{2},−\dfrac{7\pi}{iv}\correct)\) will coincide with the original solution of \(\left(3\sqrt{2}, \dfrac{\pi}{iv}\right)\). The point \(\left(−three\sqrt{2}, \dfrac{5\pi}{four}\correct)\) indicates a motion further counterclockwise by \(\pi\), which is directly opposite \(\dfrac{\pi}{4}\). The radius is expressed equally \(−3\sqrt{2}\). However, the bending \(\dfrac{5\pi}{4}\) is located in the tertiary quadrant and, as \(r\) is negative, we extend the directed line segment in the opposite direction, into the first quadrant. This is the aforementioned point every bit \(\left(three\sqrt{two}, \dfrac{\pi}{4}\right)\). The point \(\left(three\sqrt{2}, −\dfrac{7\pi}{iv}\right)\) is a move further clockwise by \(−\dfrac{7\pi}{4}\), from \(\dfrac{\pi}{four}\). The radius, \(3\sqrt{2}\), is the aforementioned.

Transforming Equations between Polar and Rectangular Forms

Nosotros can now catechumen coordinates betwixt polar and rectangular form. Converting equations can exist more than difficult, but it tin can be benign to be able to convert between the 2 forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian grade, and vice versa, we can employ the aforementioned procedures we used to catechumen points betwixt the coordinate systems. We can and so use a graphing calculator to graph either the rectangular course or the polar course of the equation.

How to: Given an equation in polar form, graph it using a graphing estimator

1. Modify the MODE to POL, representing polar form.
2. Press the Y= button to bring up a screen allowing the input of six equations: \(r_1\), \(r_2\),..., \(r_6\).
3. Enter the polar equation, set equal to \(r\).
4. Press GRAPH.

Case \(\PageIndex{5A}\): Writing a Cartesian Equation in Polar Form

Write the Cartesian equation \(x^2+y^2=ix\) in polar form.

Solution

The goal is to eliminate \(x\) and \(y\) from the equation and innovate \(r\) and \(\theta\). Ideally, we would write the equation \(r\) every bit a role of \(\theta\). To obtain the polar form, we will apply the relationships between \((x,y)\) and \((r,\theta)\). Since \(ten=r \cos \theta\) and \(y=r \sin \theta\), nosotros can substitute and solve for \(r\).

\(\begin{align*} {(r \cos \theta)}^2+{(r \sin \theta)}^2&= 9\\ r^2 {\cos}^2 \theta+r^2 {\sin}^two \theta&= nine\\ r^two({\cos}^two \theta+{\sin}^2 \theta)&= 9\\ r^2(one)&= nine\qquad \text {Substitute } {\cos}^2 \theta+{\sin}^two \theta=1\\ r&= \pm 3\qquad \text {Use the square root property.} \stop{align*}\)

Thus, \(ten^two+y^ii=9\),\(r=3\),and \(r=−3\) should generate the same graph. Come across Figure \(\PageIndex{12}\).

Figure \(\PageIndex{12}\): (a) Cartesian grade \(x^2+y^2=9\) (b) Polar form \(r=3\)

To graph a circle in rectangular form, we must offset solve for \(y\).

\[\begin{marshal*} x^ii+y^2&= 9\\ y^2&= 9-x^two\\ y&= \pm \sqrt{9-x^2} \finish{align*}\]

Note that this is two carve up functions, since a circle fails the vertical line examination. Therefore, we need to enter the positive and negative foursquare roots into the calculator separately, as two equations in the form \(Y_1=\sqrt{9−10^ii}\) and \(Y_2=−\sqrt{9−10^two}\). Press GRAPH.

Example \(\PageIndex{5B}\): Rewriting a Cartesian Equation equally a Polar Equation

Rewrite the Cartesian equation \(ten^2+y^two=6y\) as a polar equation.

Solution

This equation appears like to the previous example, just it requires different steps to convert the equation.

We tin can yet follow the same procedures we have already learned and make the following substitutions:

\(\begin{array}{ll} r^two=6y & \text{Utilise }10^2+y^two=r^2. \\ r^2=6r \sin \theta & \text{Substitute }y=r \sin \theta. \\ r^two−6r \sin \theta=0 & \text{Fix equal to }0. \\ r(r−6 \sin \theta)=0 & \text{Gene and solve.} \\ r=0 & \text{We reject }r=0 \text{, as it only represents one point, }(0,0). \\ \text{or }r=6 \sin \theta \end{array}\)

Therefore, the equations \(x^two+y^2=6y\) and \(r=6 \sin \theta\) should give us the same graph. Run across Effigy \(\PageIndex{thirteen}\).

Figure \(\PageIndex{xiii}\): (a) Cartesian course \(x^two+y^2=6y\) (b) polar form \(r=6 \sin \theta\)

The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar filigree. Clearly, the graphs are identical.

Do \(\PageIndex{4A}\):

Rewriting a Cartesian Equation in Polar Form

Rewrite the Cartesian equation \(y=3x+2\) every bit a polar equation.

Answer

We will use the relationships \(x=r \cos \theta\) and \(y=r \sin \theta\).

\(\begin{array}{cl} y=3x+2 \\ r \sin \theta=3r \cos \theta+2 \\ r \sin \theta−3r \cos \theta=2 \\ r(\sin \theta−3 \cos \theta)=ii & \text{Isolate }r. \\ r=2 \sin \theta−3\cos \theta & \text{Solve for }r. \end{array}\)

Exercise \(\PageIndex{4B}\):

Rewrite the Cartesian equation \(y^two=3−x^2\) in polar course.

Answer

\(r=\sqrt{3}\)

Identify and Graph Polar Equations by Converting to Rectangular Equations

Nosotros have learned how to convert rectangular coordinates to polar coordinates, and nosotros have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while fatigued on unlike grids, are identical.

Example \(\PageIndex{6A}\): Graphing a Polar Equation by Converting to a Rectangular Equation

Covert the polar equation \(r=2 \sec \theta\) to a rectangular equation, and depict its respective graph.

Solution

The conversion is

\[\begin{align*} r &=2 \sec \theta \\ r &= \dfrac{two}{\cos \theta} \\ r \cos \theta &=2 \\ 10 &=2 \end{marshal*}\]

Notice that the equation \(r=2 \sec \theta\) drawn on the polar grid is clearly the same every bit the vertical line \(ten=ii\) drawn on the rectangular grid (run across Figure \(\PageIndex{xiv}\)). But as \(x=c\) is the standard form for a vertical line in rectangular course, \(r=c \sec \theta\) is the standard form for a vertical line in polar form.

Figure \(\PageIndex{14}\): (a) Polar filigree (b) Rectangular coordinate system

A like give-and-take would demonstrate that the graph of the function \(r=2 \csc \theta\) will be the horizontal line \(y=2\). In fact, \(r=c \csc \theta\) is the standard form for a horizontal line in polar course, corresponding to the rectangular form \(y=c\).

Case \(\PageIndex{6B}\): Rewriting a Polar Equation in Cartesian Course

Rewrite the polar equation \(r=\dfrac{three}{1−2 \cos \theta}\) as a Cartesian equation.

Solution

The goal is to eliminate \(\theta\) and \(r\),and introduce \(10\) and \(y\). Nosotros articulate the fraction, and and then use substitution. In order to replace \(r\) with \(x\) and \(y\), we must use the expression \(x^2+y^2=r^2\).

\(\begin{assortment} r =\dfrac{3}{1−ii \cos \theta} \\ r(1−2 \cos \theta)=3 \\ r\left(1−2\left(\dfrac{x}{r}\right)\right)=3 & \text{Apply }\cos \theta=\dfrac{x}{r} \text{ to eliminate }\theta. \\ r−2x=3 \\ r=iii+2x & \text{Isolate }r. \\ r^2={(3+2x)}^two & \text{Foursquare both sides.} \\ 10^2+y^2={(three+2x)}^2 & \text{Employ }x^2+y^two=r^ii. \end{assortment}\)

The Cartesian equation is \(x^ii+y^ii={(iii+2x)}^2\). Withal, to graph it, especially using a graphing calculator or computer programme, we want to isolate \(y\).

\[\begin{align*} x^two+y^2 &= {(iii+2x)}^2 \\ y^2 &= {(iii+2x)}^2-x^ii \\ y &= \pm {(iii+2x)}^2-ten^2 \end{align*}\]

When our entire equation has been changed from \(r\) and \(\theta\) to \(10\) and \(y\), nosotros can terminate, unless asked to solve for \(y\) or simplify. Come across Effigy \(\PageIndex{15}\).

Effigy \(\PageIndex{15}\)

The "hr-glass" shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry.

Analysis

In this example, the right side of the equation tin be expanded and the equation simplified further, equally shown in a higher place. Nevertheless, the equation cannot exist written equally a single function in Cartesian form. Nosotros may wish to write the rectangular equation in the hyperbola's standard course. To do this, nosotros can start with the initial equation.

\(\brainstorm{array}{ll} ten^2+y^2={(three+2x)}^two \\ x^2+y^2−{(3+2x)}^two=0 \\ x^2+y^2−(ix+12x+4x^2)=0 \\ x^2+y^2−9−12x−4x^2=0 \\ −3x^2−12x+y^2=9 & \text{Multiply through past }−one. \\ 3x^2+12x−y^two=−9 \\ three(x^ii+4x)−y2=−9 & \text{Organize terms to consummate the square for }x. \\ 3(x^2+4x+4)−y^2=−ix+12 \\ 3{(x+2)}^2−y^two=3 \\ {(x+two)}^ii−\dfrac{y^ii}{3}=1\cease{assortment}\)

Exercise \(\PageIndex{five}\)

Rewrite the polar equation \(r=2 \sin \theta\) in Cartesian course.

Answer

\(10^2+y^two=2y\) or, in the standard class for a circumvolve, \(ten^two+{(y−one)}^two=1\)

Example \(\PageIndex{seven}\): Rewriting a Polar Equation in Cartesian Form

Rewrite the polar equation \(r=\sin(ii\theta)\) in Cartesian form.

Solution

\(\begin{array}{cl} r=\sin(2\theta) & \text{Employ the double angle identity for sine.} \\ r=2 \sin \theta \cos \theta & \text{Use }\cos \theta=\dfrac{10}{r} \text{ and } \sin \theta=\dfrac{y}{r}. \\ r=2 \dfrac{ten}{r})(\dfrac{y}{r}) & \text{ Simplify.} \\ r=\dfrac{2xy}{r^2} & \text{Multiply both sides past }r^2. \\ r^three=2xy \\ {(x^ii+y^2)}^iii=2xy & \text{As }x^2+y^two=r^2, r=\sqrt{x^2+y^2}. \end{array}\)

This equation can besides be written as

\({(x^2+y^ii)}^{\frac{three}{2}}=2xy \text{ or }ten^two+y^two={(2xy)}^{\frac{ii}{3}}\)

Key Equations

Conversion formulas

\(\cos \theta=\dfrac{x}{r} \rightarrow x=r \cos\theta\) \(\sin \theta=\dfrac{y}{r} \rightarrow y=r \sin \theta\) \(r^2=x^ii+y^ii\) \(\tan \theta=\dfrac{y}{x}\)

Key Concepts

• The polar grid is represented as a series of concentric circles radiating out from the pole, or origin.
• To plot a point in the course \((r,\theta)\), \(\theta>0\), move in a counterclockwise direction from the polar axis by an angle of \(\theta\), so extend a directed line segment from the pole the length of \(r\) in the management of \(\theta\). If \(\theta\) is negative, move in a clockwise direction, and extend a directed line segment the length of \(r\) in the direction of \(\theta\). See Example \(\PageIndex{1}\).
• If \(r\) is negative, extend the directed line segment in the contrary management of \(\theta\). Come across Example \(\PageIndex{2}\).
• To convert from polar coordinates to rectangular coordinates, use the formulas \(x=r \cos \theta\) and \(y=r \sin \theta\). Come across Example \(\PageIndex{3}\) and Example \(\PageIndex{four}\).
• To catechumen from rectangular coordinates to polar coordinates, use i or more of the formulas: \(\cos \theta=\dfrac{x}{r}\), \(\sin \theta=\dfrac{y}{r}\), \(\tan \theta=\dfrac{y}{ten}\), and \(r=\sqrt{10^two+y^two}\). Meet Example \(\PageIndex{five}\).
• Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. Come across Case \(\PageIndex{6}\), Example \(\PageIndex{7}\), and Example \(\PageIndex{8}\).
• Using the advisable substitutions makes information technology possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular aeroplane. See Case \(\PageIndex{9}\), Example \(\PageIndex{10}\), and Example \(\PageIndex{11}\).

Source: https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Algebra_and_Trigonometry_(OpenStax)/10%3A_Further_Applications_of_Trigonometry/10.03%3A_Polar_Coordinates#:~:text=To%20convert%20from%20polar%20coordinates,and%20y%3Drsin%CE%B8.

Posted by: pilgrimanable.blogspot.com

Share this post